# HW1 - 1 c 2 and c 3 such that c 1 u c 2 v c 3 w =(2 0 4 u...

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V 3.1 – 4a: Find a nonzero vector u with initial point P (-1, 3, -5) such that a) u has the same direction as v = (6, 7, -3) a) u can be any vector having P as initial point and having a terminal point, say P’ (x, y, z), such that PP’ = au where a is any positive value, not necessarily integer. For example if we pick a = 1, then PP’ = (x, y, z) – (-1, 3, -5) = (6, 7, -3). In that case P’ is (5, 10, -8) would yield a vector PP’ equal to vector u. We can pick any other positive value for “a” and end up with a different vector having the same direction as u. V 3.1 – 8: Let u, v, and w be the vectors in Exercise 6. Find scalars c

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Unformatted text preview: 1 , c 2 , and c 3 such that c 1 u + c 2 v + c 3 w = (2, 0, 4) u = (-3, 1, 2), v = (4, 0, -8), and w = (6, -1, -4) This means that we have 3 equations in 3 unknowns: Equation 1: -3c 1 + 4c 2 + 6c 3 = 2 Equation 2:c 1- c 3 = 0 Equation 3: 2c 1 – 8c 2 – 4c 3 = 4 From Equation 2: c 1 = c 3 , substituting we get: Equation 1: 3c 1 + 4 c 2 = 2 Equation 3: -2c 1 – 8c 2 = 4, or - c 1 – 4c 2 = 2. Adding the two equations, we get: 2c 1 = 4, c 1 = c 3 = 2. Subs. In Equation 1: 6 + 4c 2 = 2, 4c 2 = -4, c 2 = -1. Answer: c 1 = 2, c 2 = -1, and c 3 = 2...
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